Find all kinds of function $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfies $|x|f(y)+yf(x)=f(xy)+f(x^2)+f(f(y))$

Question

Find all kinds of function $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfies $$|x|f(y)+yf(x)=f(xy)+f(x^2)+f(f(y))$$

My attempt: \begin{align} &P(0, 0): 0 = 2f(0)+f(f(0)). \\ &\text{Let } f(0)=c \Rightarrow f(c)=-2c. \\ \ \\ &\text{Let }t > 0, \\ &P(t, 0): tf(0)=f(0)+f(t^2)+f(f(0)). \\ &\Rightarrow (t-1)c=f(t^2)-2c. \\ &\therefore (t+1)c=f(t^2). \\ &t \neq -1(\because t>0) \Rightarrow f(t)=(\sqrt{t}+1)c. \\ \ \\ &\text{Actually, it makes sense if } t=0. \\ \end{align}

Answer 1

$$|x|f(y)+yf(x)=f(xy)+f(x^2)+f(f(y)) $$

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• $(x,y)\equiv (1,1)$ $$f(f(1))=0$$
• $(x,y)\equiv (0,1)$ $$f(0)=0$$
• $(x,y)\equiv (0,-1)$ $$f(f(-1))=0$$
• $(x,y)\equiv (\sqrt{x},0)$ $$f(x)=0\;,\;\forall x\ge 0$$
• $(x,y)\equiv (-x,-1)$ $$f(-x)=xf(-1)\;,\;\forall x\ge 0$$

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$$\therefore \;\;f(x)=\begin{cases} 0&\forall x\geq 0\\ -xf(-1)&\forall x\le 0 \end{cases}$$ where $f(-1)\ge 0.$

Answer 2

Hints: You have shown that $f(t)=(1+\sqrt {|t|}) c$ for all $t\geq 0$.

To find $f(x)$ for $x <0$ put $y=-1$. The equation becomes $-xf(-1)-f(x)=f(-x)+f(x^{2})+f(f(-1))$ Since we already know the form of $f(-x)$ and $f(x^{2})$ we can find $f(x)$ from this equation. I will let you finish from here.

Answer 3

The functions are $f(x)=k(x-|x|), \; k \le 0$. The proof is listed below:
$P(0,y):yf(0)=f(f(y)) \\ P(1,1):2f(1)=2f(1)+f(f(1)) \Rightarrow f(f(1))=1 \cdot f(0)=0 \Rightarrow f(0)=0=yf(0)=f(f(y)) \\ P(x,0):0=f(x^2) \Rightarrow f(x)=0 \; \forall x>0 \\ P(-1,y) \wedge y>0:yf(-1)=f(-y) \Rightarrow f(x)=-f(-1)x \; \forall x<0$
If $f(-1)<0$, then $f(f(-1))=-f(-1)^2 \ne 0$, a contradiction, so $f(-1) \ge 0$.